Question

A small block on a frictionless horizontal surface has a mass of 2.95×10⁻² kg . It is attached to a massless cord passing through a hole in the surface. (See the figure below (Figure 1) .) The block is originally revolving at a distance of 0.295 m from the hole with an angular speed of 1.85 rad/s . The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.140 m . You may treat the block as a particle.

1.What is the new angular speed in rad/s?

2.Find the change in kinetic energy of the block.

3.How much work was done in pulling the cord?

Answer 1

Given that

A small block on a frictionless horizontal surface has a mass of (m) = 2.95×10⁻²kg

The block is originally revolving at a distance of (R1) = 0.295m

The angular speed of (w1) = 1.85 rad/s

Shortening the radius of the circle in which the block revolves to (R2) = 0.140 m

a)

Here there is no external torque acting in the system. Thus we can use conservation of angualr momentum

Now the angualr momentum for the bigger circle is

L1 =mR₁²w₁

and for smaller circle

L2 =mR₂²w2

Now the new angular speed is

w2 =(R1/R2)²* w1 = (0.295/0.140)²*1.85 =8.214rad/s

b)

The change in kinetic energy is =(1/2)(I₂w₂²-I₁w₁²)

=(1/2)[mR₂²w2²-mR₁²w1²)

   =(2.95×10⁻² kg/2)[(0.140*8.214)²-(0.295*1.85)²]

   =1.475*10-2[1.322-0.2978]=1.510*10^-2J

c)

We know that from the work energy theorem the work done is equal to the change in kinetic energy

W =Change in kinetic energy =DeltaKE =1.510*10^-2J